Class 4: More Pendulum results

Class 4: More Pendulum results The pendulum is a mechanical problem with a long and interesting history, from Galileo s first ansatz that the period was independent of the amplitude based on watching priests swinging incense burners, to the modern study of chaotic systems, the pendulum has been there. Pendulum clocks became the very model of deterministic science after Newton, when analogies were made to a clockwork universe that obeyed the laws of physics and was completely determined by its initial conditions. From a paradigm of determinism to a classic example of unpredictable chaos, the pendulum is an ideal system to learn some basic dynamics. We calculated the formal analytic solution last time and found that, although useful for proving various properties of pendulum dynamics, we need to the numerical solution to build physical intuition for the system. Pendulum orbits: Numerical solution Time series of angle and angular velocity We plotted the analytic solution last time and, at least in one situation, we got gaps in the plot, probably due to insufficient sampling (or possibly no-unique behavior of the Elliptic functions in some regions). Here's a numerical solution of the pendulum equation of motion θθ + ωω 0 sinθθ = 0 with ωω 0 =. θ0 =.5; s = NDSolve[{θ''[t] + Sin[θ[t]] == 0, θ[0] θ0, θ'[0] == 0}, θ, {t, 0, 30}]; Plot {Evaluate[{θ[t]} /. s], θ0 Cos[t]}, {t, 0, 30}, PlotStyle {Thick, Automatic}, Frame True, FrameLabel {"t", "θ"}, RotateLabel False, AspectRatio GoldenRatio.5.0 0.5 θθ 0.0 0.5.0.5 0 5 0 5 0 5 30 t The blue curve is the actual pendulum - the Jacobi elliptic function - and the orange curve is the cosine (harmonic oscillator) oscillation. You can see that the shape of the curves are similar but that the frequencies are different. In fact, nonlinear oscillators like the pendulum usually have amplitudedependent frequencies or frequency-dependent amplitudes, depending on how you look at it. This

dependent frequencies or frequency-dependent amplitudes, depending on how you look at it. This plot is for an initial displacement angle of θθ 0 =.5 radians, almost 90 degrees. If we increase the initial angle to 3 radians (about 7 degrees) we get: θ0 = 3.0; s = NDSolve[{θ''[t] + Sin[θ[t]] == 0, θ[0] θ0, θ'[0] == 0}, θ, {t, 0, 50}]; Plot {Evaluate[{θ[t]} /. s], θ0 Cos[t]}, {t, 0, 50}, PlotStyle {Thick, Automatic}, Frame True, FrameLabel {"t", "θ"}, RotateLabel False, AspectRatio GoldenRatio 3 θθ 0 3 0 0 0 30 40 50 t This is the case that gave problems plotting from the analytic solution. No problem numerically. For this larger amplitude the harmonic oscillator curve has the same frequency it did for the smaller amplitude, but the pendulum frequency is now much lower. The shape of the pendulum curve even looks different than the cosine function -- it's curvature at the maxima no longer looks sinusoidal, but is more rounded. You're seeing the difference between a harmonic (linear) oscillation and one kind of nonlinear oscillation. Internalize it. Before moving on, it's worth looking at the pendulum motion for a small initial angle, say 0.07 radians, or about. θ0 = 0.07; s = NDSolve[{θ''[t] + Sin[θ[t]] == 0, θ[0] θ0, θ'[0] == 0}, θ, {t, 0, 30}]; Plot {Evaluate[{θ[t]} /. s], θ0 Cos[t]}, {t, 0, 30}, PlotStyle {{Thick, Dashed}, Automatic}, Frame True, FrameLabel {"t", "θ"}, RotateLabel False, AspectRatio GoldenRatio For this small angle, the pendulum motion is virtually identical to the cosine function (yes, there are curves plotted here, they just overlap and look like one). You can easily verify this theoretically, by solving the equation of motion (4) for small angles (use the first term in a Taylor series for the sine function) - you'll find the small angle oscillation is sinusoidal with fixed frequency ωω 0. Producing a Phase Portrait for the pendulum Phase space plots, like the one we obtained last class, show us how the dynamical variables are related. But, for the pendulum, we already know how they are related: by the energy integral! We don't need a numerical solution to get a phase plot, just use the conservation of energy equation : m L θθ + m ωω0 L cosθθ = E

3 Solving this equation for θθ + gives a relation that represents a curve in θθ-θθ + space. Each curve will be associated with a different value of the total energy E. Before we plot this curve, let's try to use our physical intuition to figure out what we expect the result to look like. For example, for small angles, θθ <<, we know the pendulum motion will be a simple oscillation - i.e. swinging back and forth - also called libration. If we start the pendulum with zero angular velocity and initial angle θθ 0 we expect it to traverse a curve not unlike the phase space orbit above: it will move downward and to the left until it gets to θθ = 0, where its angular velocity will reach a maximum. After that it will swing over to negative angles until it reaches its maximum angle θθ 0 with zero θθ +, then returning to θθ = 0 with a positive θθ +, and finally reaching its initial position with zero angular velocity - only to repeat the whole operation ad infinitum. On a θθ-θθ + graph, this would look like a closed curve. Let's let Mathematica plot it exactly for us: define En = E / m L so, and let's plot for a sample frequency ωω 0 =, so the energy integral becomes θθ + cosθθ = E m L = En, or, converting to Mathematica language: θdot = Cos[θ] + En, Cos[θ] + En ; As noted above, small oscillations will have negative energies, so let's choose En = 0.5 for an example: c = Plot θdot /. En.5, {θ, π /, π / }, PlotPoints 00, Frame True, FrameLabel "θ", "θ ", RotateLabel False, AspectRatio GoldenRatio.0 0.5 θθ + 0.0 0.5.0.5.0 0.5 0.0 0.5.0.5 θθ This nearly circular phase space orbit for E / m L = 0.5 has an amplitude of (can you prove this?). Now, imagine a different initial condition, say starting the pendulum bob at θθ 0 = ππ, with positive initial velocity θθ + 0 > 0. What happens? A little thought should tell you that the pendulum would now rotate, since it has kinetic energy at the top of its swing so when it reaches that maximum height again, it retains that energy and will keep moving. What does this rotation look like on a phase-space plot? It would start at θθ = ππ then increase upward and leftward until it reaches θθ = 0, again with a maximum angular velocity. Then it would swing into the negative angle range until it gets to θθ = ππ again, where it would have angular velocity equal to its initial velocity - its minimum angular speed - and then it would keep going, repeating this pattern for larger and larger negative angles. In Mathematica, it would look like this:

4 c = Plot θdot /. En 3, {θ, 5 π, π}, PlotPoints 00, PlotStyle {Purple}, Frame True, FrameLabel "θ", "θ ", RotateLabel False, AspectRatio GoldenRatio 3 θθ + 0 3 5 0 5 0 θθ The upper curve is the one we just described, starting with a positive angular velocity. The lower curve is the corresponding trajectory for a negative initial angular velocity at the same energy. Combining these plots for two different initial conditions on the same axes gives: Show[c, c, PlotRange All] 3 θθ + 0 3 5 0 5 0 θθ So, we can easily see the difference between libration (swinging) and rotation by looking at the graph. In fact, we can guess that there should be a boundary curve that divides regions of libration from regions of rotation: a curve that's usually called the separatrix. Here it is for the example case we're plotting:

5 c3 = Plot θdot /. En, {θ, 5 π, π}, PlotPoints 00, PlotStyle {Red}, Frame True, FrameLabel "θ", "θ ", RotateLabel False, AspectRatio GoldenRatio ; Show[c, c, c3, PlotRange All] 3 θθ + 0 3 5 0 5 0 θθ What does the separatrix curve represent physically for the pendulum? It's an idealized orbit that can't really be achieved with a real pendulum: for example, start it at the top of its swing, θθ = ππ, with zero velocity - if it could then somehow be infinitesimally perturbed (with no change in energy) so it would swing, it would swing around and come back to the top and stop (and it would take an infinite time to do it)! That's the separatrix orbit. These three orbit types are characteristic of the pendulum - all orbits of a plane pendulum are of one of these types. Let's plot some other orbits (i.e. different energies) to get a global picture of pendulum behavior.

6 θdotp = θdot /.En.5; θdotp = θdot /.En.; θdotp3 = θdot /.En ; θdotp4 = θdot /.En ; θdotp5 = θdot /.En 3; θdotp6 = θdot /.En 4; Plot {θdotp, θdotp, θdotp3, θdotp4, θdotp5, θdotp6}, {θ, 3 π, 3 π}, Frame True, PlotPoints 00, FrameLabel "θ", "θ ", RotateLabel False, PlotLabel "Pendulum Phase Portrait" 3 Pendulum Phase Portrait θθ + 0 3 5 0 5 θθ This plot is called a phase portrait for the pendulum. Does this agree with your calculation of the turning points in Homework? From the phase portrait you can determine all the possible types of behavior of this system with two variables (θθ and θθ + ). Phase portraits can be an extremely useful tool for initial investigations of a new nonlinear system. Note that the phase portrait is really just a way of visualizing the energy integral (each curve is a solution to the energy integral with a different constant energy), so we see that the energy integral really contains all the information we need to understand the motion! Using a contour plot to visualize the Phase Portrait for the pendulum Here's another, probably easier, way of viewing the phase portrait: the energy integral can be thought of as a two-dimensional function, energy E θθ, θθ + as a function of the two dynamical variables, θθ and θθ +. If you plot a contour map of the energy function, the contours will be at constant values of energy and therefore will represent the phase space orbits at each energy. Rewriting the energy integral:

7 energy = θdot Cos[θ]; ContourPlot[energy, {θ, 0, 0}, {θdot, 3.5, 3.5}, ImageSize Small] This default plot can be made more useful with options. For example, let's get rid of the color shading between contours. Also, one advantage of the contour plot is that we can label each orbit (contour line) with its energy: ContourPlot energy, {θ, 0, 0}, {θdot, 3.5, 3.5}, ContourShading False, ContourLabels All, AspectRatio Automatic, FrameLabel "θ", "θ ", RotateLabel False, PlotLabel "Pendulum Phase Portrait" If you don't like the default contour energy values, you can select your own. If you want tweak further, you can specify the significant figures and number of decimal places for the labels. Let s also clean up the plot by eliminating those gaps at multiples of θθ = ππ, and by labeling the θθ-axis in multiples of ππ:

8 θmin = 0; θmax = 0; tickmin = Ceiling θmin π ; tickmax = Floor[θmax / π]; pendpp = ContourPlot energy, {θ, θmin, θmax}, {θdot, 3.5, 3.5}, PlotPoints 50, ContourShading False, Contours {0.53, 0.0,.0,.5, 5.0}, ContourStyle Blue, ContourLabels (Text[NumberForm[#3, {, }], {#, #}] &), FrameTicks {Table[tickmin π + k π, {k, 0, tickmax tickmin}], Automatic}, AspectRatio Automatic, FrameLabel "θ", "θ ", RotateLabel False, PlotLabel "Pendulum Phase Portrait", ImageSize Large The NumberForm command tells Mathematica to label variable 3 (energy) with significant figures and decimal places. I also added more sampling points with the PlotPoints command to clear up that "gap" in the separatrix curve; the extra sampling looks nicer, but does take more compute time. Finally, I changed the θθ axis to be in multiples of ππ radians using the Ticks command. Compare the pendulum phase portrait to that of the simple harmonic oscillator I mentioned above that if you approximate the sine function near θθ = 0 with the first term in its Taylor series, that you end up with the equation of motion for a simple harmonic oscillator: θθ + ωω 0 θθ 0 which (as you no doubt learned in PHY 0...) has a simple sinusoidal solution θθ(t) = A sinωω 0 t + B cosωω 0 t. It's not hard to show that the energy integral for this equation is θθ+ + ωω 0 θθ = E m L Let's plot the harmonic oscillator phase portrait using this energy equation. Calling energy = E / m L and taking ωω 0 = :

9 energysho = θdot + θ ; shopp = ContourPlot energysho, {θ, 4, 4}, {θdot, 3.5, 3.5}, PlotPoints 00, ContourShading False, Contours {0.0,.0,.5, 5.0, 7.5}, ContourStyle Red, AspectRatio Automatic, FrameLabel "θ", "θ ", RotateLabel False, PlotLabel "Simple Harmonic Oscillator Phase Portrait" ; Show[pendPP, shopp, PlotRange {{4, 4}, Automatic}, PlotLabel "Pendulum (blue), Harmonic oscillator (red)"] You can see that the oscillator phase portrait is just a set of concentric circles representing oscillations: no rotational motion and no separatrix. Thus it is only a good approximation for very small angles, that is it's valid locally, near the origin (0, 0) of the phase plot. Using the small angle Taylor expansion to first order is sometimes called linearizing the equation of motion locally. The global behavior, shown in the full pendulum phase plot is quite different. As we shall see, we can often learn some important information by linearizing locally, but we need to be careful that we don't extrapolate the local behavior to the global system, which may be totally different.

0 Higher order approximation to the pendulum equation You looked at the third order approximation to the pendulum equation in Class. Now lets look at higher order approximations. 3th order approximation Here s an example of the 3 th order approximation to the pendulum force (keeping 3 terms in the Taylor series): order3 = Normal[Series[Sin[θ[t]], {θ[t], 0, 3}]] EOM = θ''[t] + ω0 (order3) En = Integrate[EOM θ'[t], t] θ[t] θ[t]3 6 + θ[t]5 0 θ[t]7 5040 + θ[t]9 θ[t] 36 880 39 96 800 + θ[t]3 6 7 00 800 ω0 θ[t] θ[t]3 6 + θ[t]5 0 ω0 θ[t] 4 ω0 θ[t] 4 + 70 ω0 θ[t] 6 ω0 θ[t] 8 40 30 + ω0 θ[t]0 3 68 800 θ[t]7 5040 + θ[t]9 θ[t] 36 880 39 96 800 + θ[t]3 6 7 00 800 + θ [t] ω0 θ[t] ω0 θ[t]4 + 479 00 600 87 78 9 00 + θ [t] Prepare to plot by getting separatrix locations. One way to do that is to locate the fixed points of the motion: where the acceleration and the velocity vanish. The acceleration here is proportional to the series for sinθθ, i.e. what I called oder3 above. Let s set that equal zero and find the real solutions only: Solve[order3 0, θ[t], Reals]; N[%] {{θ[t] 0.}, {θ[t] 5.97835}, {θ[t] 3.46}, {θ[t] 3.46}, {θ[t] 5.97835}} We now have positive real roots (and two negative). Get the energy of the first real root: En = En /. ω0 /. {θ[t] θ, θ'[t] ω} Es = En /. {ω 0, θ 3.46} θ. θ4 4 + θ6 θ8 70 40 30 + θ0 θ 3 68 800 479 00 600 + θ4 87 78 9 00 + ω Get the energy of the second real root: Es = En /. {ω 0, θ 5.97835} 0.59794 Here are the plots:

approx3 = ContourPlot En, {θ, 8, 8}, {ω, 3, 3}, PlotPoints 50, ContourShading False, Contours {0.0, Es, 0.50,.00,.50, Es,.50, 3.0, 3.5, 4.0, 4.5, 5.0}, ContourStyle Blue, FrameLabel θ, θ, RotateLabel False, AspectRatio Automatic, ImageSize Large ; Epend = ω + ( Cos[θ]); pend = ContourPlot Epend, {θ, 8, 8}, {ω, 3, 3}, PlotPoints 50, ContourShading False, Contours {0.0, Es, 0.50,.00,.50, Es,.50, 3.0, 3.5, 4.0, 4.5, 5.0}, ContourLabels True, ContourStyle Red, FrameLabel θ, θ, RotateLabel False, AspectRatio Automatic, ImageSize Large ; Show[approx3, pend, PlotRange All, PlotLabel "Full pendulum (red) and 3th order (blue)"] We get pretty good agreement out to an angle of about 5 radians, but the separatrix structure is still inaccurate. As mentioned in class, note that for every other order (5 th, 9 th, 3 th, etc.) the unphysical orbits that blow up are not present. 5th order approximation Here s an example of the 5 th order approximation (keeping 5 terms in the Taylor series): order5 = Normal[Series[Sin[θ[t]], {θ[t], 0, 5}]] EOM = θ''[t] + ω0 (order5); En = Integrate[EOM θ'[t], t]; θ[t] θ[t]3 6 + θ[t]5 0 θ[t]7 5040 + θ[t]9 θ[t] 36 880 39 96 800 + θ[t]3 θ[t]5 6 7 00 800 307 674 368 000 Prepare to plot by getting fixed point locations (real solutions only) & real root energies:

Solve[order5 0, θ[t], Reals]; N[%] En = En /. ω0 /. {θ[t] θ, θ'[t] ω}; Es = En /. {ω 0, θ 3.459} Es = En /. {ω 0, θ 6.4605} {{θ[t] 0.}, {θ[t] 7.057}, {θ[t] 6.4605}, {θ[t] 3.459}, {θ[t] 3.459}, {θ[t] 6.4605}, {θ[t] 7.057}}. 0.038949 Here are the plots: approx5 = ContourPlot En, {θ, 8, 8}, {ω, 3, 3}, PlotPoints 50, ContourShading False, Contours {Es, 0.0, 0.06, 0.50,.00,.50, Es,.50, 3.0, 3.5, 4.0, 4.5, 5.0}, ContourStyle Blue, FrameLabel θ, θ, RotateLabel False, AspectRatio Automatic, ImageSize Large ; pend = ContourPlot Epend, {θ, 8, 8}, {ω, 3, 3}, PlotPoints 50, ContourShading False, Contours {Es, 0.0, 0.06, 0.50,.00,.50, Es,.50, 3.0, 3.5, 4.0, 4.5, 5.0}, ContourLabels True, ContourStyle Red, FrameLabel θ, θ, RotateLabel False, AspectRatio Automatic, ImageSize Large, PlotLabel "Full Pendulum: phase portrait" ; Show[approx5, pend, PlotRange All, PlotLabel "Full pendulum (red) and 5th order approximation (blue)"] For this approximation the infinite orbits return at right and left. We get pretty good agreement out to an angle of a little over 5 radians, and now there are two unstable positive fixed points, but the second separatrix is not connected to the first! 7th order approximation Here s an example of the 7 th order approximation (keeping 7 terms in the Taylor series):c

3 order7 = Normal[Series[Sin[θ[t]], {θ[t], 0, 7}]] EOM = θ''[t] + ω0 (order7); En = Integrate[EOM θ'[t], t]; Solve[order7 0, θ[t], Reals]; N[%] En = En /. ω0 /. {θ[t] θ, θ'[t] ω}; Es = En /. {ω 0, θ 3.459}; Es = En /. {ω 0, θ 6.839}; Es3 = En /. {ω 0, θ 9.494}; θ[t] θ[t]3 + θ[t]5 6 0 θ[t] 5 307 674 368 000 + θ[t]7 355 687 48 096 000 θ[t] 5 090 94 7 709 440 000 θ[t] 5 5 5 0 043 330 985 984 000 000 θ[t]7 5040 + θ[t]9 θ[t] 36 880 39 96 800 + θ[t]3 6 7 00 800 θ[t]9 645 00 408 83 000 + θ[t]3 5 85 06 738 884 976 640 000 + θ[t]7 0 888 869 450 48 35 60 768 000 000 {{θ[t] 0.}, {θ[t] 9.494}, {θ[t] 6.839}, {θ[t] 3.459}, {θ[t] 3.459}, {θ[t] 6.839}, {θ[t] 9.494}} approx7 = ContourPlot En, {θ, 3, 3}, {ω, 3, 3}, PlotPoints 50, ContourShading False, Contours {Es, Es, Es3, 0.50,.00,.5,.50, 3.0, 3.5, 4.0, 4.5, 5.0}, ContourStyle Blue, FrameLabel θ, θ, RotateLabel False, AspectRatio Automatic, ImageSize Large ; pend = ContourPlot Epend, {θ, 3, 3}, {ω, 3, 3}, PlotPoints 50, ContourShading False, Contours {Es, Es, Es3, 0.50,.00,.5,.50, 3.0, 3.5, 4.0, 4.5, 5.0}, ContourLabels True, ContourStyle Red, FrameLabel θ, θ, RotateLabel False, AspectRatio Automatic, ImageSize Large ; Show[approx7, pend, PlotRange All, PlotLabel "Full pendulum (red) and 7th order approximation (blue)"] OK, things are looking up! We get good agreement out to an angle of about 0 radians, including the separatrix with multiple fixed points, after which the divergent orbits show up. Conclusion: expanding the force in a Taylor series really does get more and more accurate as the order of the approximation increases -- even though the actual terms are minuscule (e.g. 7 order term has coefficient 0 8 ).

4 Fixed points of the pendulum equation The phase portrait for the pendulum has some obvious points of interest: the points where the separatrix appears to intersect itself, for example (θθ = ( n + )ππ, i.e. odd multiples of ππ) and the centers of the oscillatory orbit regions (θθ = nππ, i.e. even multiples of ππ). A few moments thought and you ll realize that if you start the pendulum at any of those points, nothing happens! They re called fixed points or equilibrium points, and are defined by zero acceleration and velocity. From the equation of motion θθ + ωω 0 sinθθ = 0 with ωω 0 =, we see that zero acceleration implies sinθθ = 0. So the fixed points occur at all angles θθ * satisfying θθ * = sin 0, or fixed point position: θθ * = n ππ, for any integer n We can determine what the motion is like near the fixed point by linearizing about the fixed point, i.e. expanding the force to first order, dropping higher order terms: F(θθ) = F(θθ * ) + F θθ * (θθ θθ* ) = F θθ * (θθ θθ* ) where the second equality is true because the Force must vanish at any fixed point. The next step is to guess an exponential solution θθ θθ * = A e αα t and solve for the constant αα: any positive real value implies the motion leaves the fixed point (motion is unstable), and imaginary αα implies oscillation about the fixed point (a type of stable motion). Let s try the pendulum fixed points: F = sinθθ (for ωω 0 = ) so the linearized equation of motion is θθ = (cosθθ) * (θθ θθ * ). Fixed point θθ * = 0: In this case, the linearized equation becomes θθ = θθ and trial solution θθ = A e αα t gives αα = so αα = ±i, which implies oscillatory motion about the fixed point, a type of stable motion. Fixed point θθ * = ππ: In this case, the linearized equation becomes θθ = +θθ and trial solution θθ = A e αα t gives αα = + so αα = ±, which implies exponential motion, with the positive exponent giving unstable motion. It should be clear that all the fixed points at multiples of ππ will be the same as the zero fixed point, and the same for thee odd multiples of ππ: they ll be the same as the θθ * = ππ fixed point. Hence: Pendulum fixed points: Fixed point θ * = n π θ * = ( n + ) π Stability Stable Unstable where n is an integer. This agrees with the phase portrait above: there are closed loop orbits surrounding the stable fixed points, implying oscillation, while most orbits leave the unstable points, implying instability. Note that you will have both oscillatory and rotating orbits in the neighborhood of the unstable fixed points.

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